Problem 29 - Calculus Volume 1

Question

A cylinder is leaking water but you are unable to determine at what rate. The cylinder has a height of \(2 \mathrm{~m}\) and a radius of \(2 \mathrm{~m}\). Find the rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is \(10 \mathrm{~cm} / \mathrm{min}\) when the height is \(1 \mathrm{~m}\).

Step-by-Step Solution

Answer

The water leaks at a rate of \(-0.4\pi \) cubic meters per minute.

Step 1: Understand the Problem

The problem involves a cylinder with a radius of 2 m and an initial height of 2 m. We need to find the rate at which water leaks when the height of the water is 1 m, given that the rate of change of the height (\( \frac{dh}{dt} \)) is decreasing at a rate of 10 cm/min.

Step 2: Convert Units

Convert the rate of height decrease to meters per minute, as it is currently in centimeters. So, \( 10 \text{ cm/min} \) is equal to \( 0.1 \text{ m/min} \).

Step 3: Write the Volume Formula

The volume \( V \) of a cylinder is given by the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. With \( r = 2 \text{ m} \), the formula becomes \( V = 4\pi h \).

Step 4: Differentiate the Volume with Respect to Time

Differentiate the volume formula with respect to time \( t \). The derivative is \( \frac{dV}{dt} = 4\pi \frac{dh}{dt} \), which relates the rate of change of volume to the rate of change of height.

Step 5: Substitute Known Values into the Equation

Now substitute \( \frac{dh}{dt} = -0.1 \text{ m/min} \) into the differentiated equation: \( \frac{dV}{dt} = 4\pi (-0.1) \).

Step 6: Calculate the Rate of Leak

Compute the expression to find \( \frac{dV}{dt} = -0.4\pi \). This gives us the rate at which water is leaking in cubic meters per minute.

Key Concepts

Cylinder Volume Calculation

To solve problems that involve cylinders, it's important to know how to calculate the cylinder's volume. A cylinder is a three-dimensional shape with two parallel bases connected by a curved surface. The distance between these bases is the height, and they both have the same radius. The formula for the volume of a cylinder is given by:\[ V = \pi r^2 h \]

• \( V \) is the volume of the cylinder,
• \( r \) is the radius of the base,
• \( h \) is the height of the cylinder,
• \( \pi \) is a constant approximately equal to 3.14159.

In our specific problem, the cylinder has a radius of 2 m. So we modify the formula to this: \[ V = 4\pi h \]This allows us to calculate the volume of water in the cylinder at any given height \( h \). Since both the radius and height of the cylinder are critical to determine its volume, knowing and applying this formula is essential.

Rate of Change

The rate of change signifies how one quantity changes in relation to another. This concept is at the heart of many calculus problems, especially those involving real-world applications like our leaky cylinder. Here, the rate of change of the height of the water in the cylinder is given as 10 cm/min. Because our volume formula uses meters, converting this rate of height decrease is necessary. Once converted to meters per minute, it becomes 0.1 m/min. This negative rate indicates a decrease in water height. It directly impacts how rapidly the volume of the cylinder changes over time. In essence, the taller the cylinder (or higher the water), the larger the volume, which means that a change in height translates directly into a change in volume as well. Understanding the rate of change helps us maintain accuracy when calculating how rapidly water is escaping the cylinder.

Differentiation with Respect to Time

Differentiation is a crucial tool in calculus for finding how a quantity changes over time. It is particularly useful when dealing with dynamic situations, such as changing volumes of liquids. When we differentiate with respect to time, we denote this change using derivatives. In our problem, we need to find out how the volume (\( V \)) of the cylinder changes with time (\( t \)). To achieve this, differentiate the volume formula with respect to \( t \):\[ \frac{dV}{dt} = 4\pi \frac{dh}{dt} \]This equation links the rate of volume change, \( \frac{dV}{dt} \), with the rate of height change, \( \frac{dh}{dt} \). After substituting the known rate of height change from our step-by-step solution (which is \(-0.1 \text{ m/min}\)), we calculate the rate of leakage of water as:\[ \frac{dV}{dt} = 4\pi (-0.1) = -0.4\pi \text{ cubic meters per minute} \]Differentiation with respect to time provides a powerful way to model and solve real-world problems like these, by quantifying the rapidity with which changes occur.